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Gibbs Free Energy (GFE) is the amount of change of energy during a reaction, either energy
consumed or generated.
The movement of any molecule (non-charged) down, or up a concentration gradient involves a
change in free energy, ΔG ("Delta G")
Moving down the concentration gradient releases energy so ΔG is negative
Moving up the concentration gradient consumes energy so ΔG is positive
The amount of free energy released or consumed can be calculated from the equation
ΔGin means the energy either released, or must be put in to the system, for the molecule to
move from the ECF into the ICF, in other words, molecule moving into the cell. That’s what
the “in” means with the term ΔGin.
What this means is that:
when [X]in is less than [X]out, then ΔG will be negative.
When [X]in is more than [X]out, then ΔG will be positive.
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Facilitated Diffusion of Glucose
Assume:
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a concentration of glucose inside the cell of 0.5 millimolar (mM/L), and a
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concentration of glucose outside the cell of 5 millimolar (mM/L)
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a body temperature of 37°C, so an absolute temperature of 37 + 273 = 310°K, and
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the plasma membrane is permeable to glucose. As I’ve stated, the plasma membrane
per se is not very permeable to glucose but in this case, we’re factoring in diffusion via
transporters as included in the concept of being “permeable”.
ΔGin = (2)(273+37) x ln (0.5/5)
= (2)(310) x ln (0.1)
= (620)(−2.3) = −1,426 cal/mole
= −1.4 kcal/mole
(If you prefer to work with log10, multiply the log10 by 2.303, thus log10 (0.1) = −1; −1 x 2.303 =
−2.3) FYI
Because the process proceeds with the release of free energy (negative GFE), it can proceed
spontaneously. In this case, facilitated diffusion would be required because glucose needs
transport channels to allow it to pass through the lipid bilayer of the plasma membrane.
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Active Transport of Glucose
Filtration of the blood in the glomeruli of the kidneys produces “filtrate” with a concentration
of glucose about the same as that of the blood (~ 5 mM/L).
All of this glucose is normally reclaimed by transport
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from the filtrate within the proximal tubule
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across the plasma membrane at the apical surface of the epithelial cells lining the tubule
and
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across the plasma membrane at the basolateral surface of the cell into the interstitial
fluid and
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back into the blood.
As the process continues and more and more glucose is removed from the filtrate, the
concentration gradient against which the glucose must be pumped by active transport,
increases. This means that GFE for glucose in this scenario must be positive.
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Problem:
What is the free energy needed to move glucose from the filtrate into the proximal tubule cells
when the concentration in the filtrate has dropped to 0.006 mM/L?
The problem is to pump glucose into the cell where it is about 6.0 mM/L. So, the total gradient
through which the glucose must be pumped is 0.006 mM/L → 6.0 mM/L.
ΔGin = (2)(310) x ln (6.0/.006)
= 620 ln (1000) = (620)(6.9) = + 4,278 cal/mole
= + 4.3 kcal/mole What this means is that more than 4.3 kcal/mole is needed to move
glucose against is gradient.
Where is the needed energy to come from?
The Na+/glucose transporter
The active transport of glucose is mediated by the Na+/glucose transporter.
This is a symporter because both the sodium ion and the glucose molecule are passing through
the membrane in the same direction:
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sodium DOWN its gradient of about
o
140 mM/L outside to
o
10 mM/L inside
while glucose is going against its gradient (0.006 mM/L → 6 mM/L).
A mole of sodium ions (Na+) moving down this concentration gradient
releases −1.6 kcal/mole of free energy.
ΔGin = (2)(310) ln (10/140)
= (620) ln (0.07) = (620)(− 2.64) = −1,637 cal/mole
= −1.6 kcal/mole
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Is this enough to move a mole of glucose?
No, but there is another force we must consider.
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Sodium ions carry a single positive charge and
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the interior of the cell is negatively charged
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So, the attraction between opposite charges provides a second force for bringing
Na+ into the cell.
This, too, can be quantified.
ΔG = (z)(F)(Vm) where
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z = the charge on the ion (+1 in this case)
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F = 23,062 = the calories released as one mole of charge moves down a voltage gradient
of 1 volt (1000 mV)
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Vm = the membrane potential, about - 70 mV or -0.07V in mammalian cells. Vm must be
in Volts
ΔG = (+1)(23,062)(- 0.07)
= −1,614 cal/mole (−1.6 kcal/mole) Adding this to the free energy available from the
concentration difference ( −1,637 cal/mole) gives us a total yield of
about -3.2 kcal/mole from this combined or electrochemical gradient.
We need more than 4.3 kcal/mole to move the glucose against its concentration gradient to
get it into the cell. So, -3.2 kcal/mole is still not enough to get the job done. But some of
these specialized Na+/glucose symporters actually use two sodium ions for every one
molecule of glucose. So, we double the GFE for sodium. 2 x -3.2 kcal/mole = -6.4 kcal/mole,
which is more than the 4.3 kcal/mole needed to move glucose up its concentration gradient.
Problem solved!
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Many cellular processes depend on a strong Na+ concentration gradient. Maintaining this
gradient is the job of the Na+/K+ ATPase pump, moving 3 Na+’s out of the cell and moving 2
K+’s into the cell, both against their concentration gradients. The energy for this comes from
ATP. The Na+/K+ ATPase pump is the greatest single user of energy in our body.
GFE and Na+/K+ ATPase pump
Suppose you had the following concentrations of Na+ and K+:
ICF - Na+: 15 mM/L, K+: 140 mM/L
ECF - Na+: 145 mM/L, K+: 5 mM/L
and suppose the resting membrane potential is -70mV.
How much energy needs to be supplied to the Na+/K+ ATPase pump
in order for it to move the Na+/K+?
Concentration calculation
For Na+: ΔGin = (2)(310) x ln (15/140)
(620)(-2.2) = -1,364 cal/mole or -1.4 kcal/mole
For K+: ΔGin = (2)(310) x ln (140/5)
(620)(3.3) = 2,046 cal/mole or 2.1 kcal/mole
Electrical force calculation
For Na+: ΔG = (+1)(23,062)(- 0.07)
= −1,614 cal/mole (−1.6 kcal/mole)
For K+: ΔG = (+1)(23,062)(- 0.07)
= −1,614 cal/mole (−1.6 kcal/mole)
Total ΔGin for Na+: -1.4 kcal/mole + -1.6 kcal/mole = -3.0 kcal/mole x 3 = -9.0 kcal/mole
Total ΔGin for K+: 2.1 kcal/mole + -1.6 kcal/mole = 0.5 kcal/mole x 2 = 1.0 kcal/mole
9.0 + 1.0 = 10 kcal/mole to move 3 Na+’s out of the cell and 2 K+’s into the cell.
Therefore, hydrolysis of ATP must yield greater than 10 kcal/mole to supply the Na+/K+ ATPase
pump.
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